Problem: Factor completely. $2x^2-162=$
Answer: First, we take a common factor of $2$. $2x^2-162=2(x^2-81)$ Now, let's factor $x^2-81$. Both $x^2$ and $81$ are perfect squares, since $x^2=({x})^2$ and $81=({9})^2$. $x^2-81 = ({x})^2-({9})^2$ So we can use the difference of squares pattern to factor. ${a}^2 - {b}^2 =({a}+{b})({a}-{b})$ In this case, ${a}={x}$ and ${b}={9}$ : $({x})^2 - ({9})^2 =({x}+{9})({x}-{9})$ $\begin{aligned} 2x^2-162&=2(x^2-81) \\\\ &=2(x+9)(x-9) \end{aligned}$ In conclusion, the complete factorization is: $2(x+9)(x-9)$ Remember that you can always check your factorization by expanding it.